Answer
0.15.
Work Step by Step
We note that 55 mph is 24.582 m/s. Calculate the energy per second (power) needed to overcome the drag forces.
$$P_{car}=Fv=(350N)( 24.582 m/s)=8604W$$
Calculate the heat energy per second (power) delivered by the gasoline. This is the energy content of a gallon of gas, divided by the time it lasts. A car that gets 32 mph while traveling at 55 mph will burn one gallon of gasoline in 32/55 hours, or 2094.45 seconds.
$$P_{gas}=\frac{(3.2\times10^7J/L)(3.8L/gallon)}{2094.45 s }=58056W$$
Equation 15-4a says that the ratio of these is the efficiency.
$$e=\frac{W}{Q_H}=\frac{W/t}{Q_H/t}=\frac{8604W}{58056}=0.148\approx 0.15$$